Formula to calculate 2cos inverse x for different x

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\[ cos^{-1}(x)=\theta \] then \[ x=cos\theta \]................................................(1)

Case 1 \[0\leqslant x\leq 1\]

 \[0\leqslant x\leq 1\]

\[0 \leq cos\theta \leq 1\]

\[0 \leq \theta \leq \frac{\Pi }{2}\]

\[0 \leq 2\theta \leq \Pi \]

Also \[0 \leq x \leq 1\]

\[-1\leq 2x^2-1 \leq 1\]

\[cos2\theta =2x^2-1 \]

\[2\theta = cos^{-1}(2x^2-1 )\]

\[2cos^{-1}(x) = cos^{-1}(2x^2-1 )\]

Case 2  \[ -1\leq x\leq 0\]

That implies \[ -1\leq cos\theta \leq 0\]              from equation 1

\[\frac{\Pi }{2}\leq \theta \le\Pi \]

\[\Pi \leq 2\theta \le2\Pi \]

\[-2\Pi \leq -2\theta \le-\Pi \]

\[0 \leq2\pi -2\theta \le\pi \]

Also \[ -1\leq x\leq 0\]

hence \[ -1\leq 2x^2-1\leq1\]

\[cos2\theta= 2x^2-1\]

\[cos(2\pi-2\theta)= 2x^2-1\]

\[(2\pi-2\theta)= cos^{-1}(2x^2-1)\]

\[ 2\pi-2cos^{-1}(x)=cos^{-1}(2x^2-1)\]

Therefore \[2cos^{-1}(x)=2\pi-cos^{-1}(2x^2-1)\]

hence proved